Chapter 15

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# Chapter 15 - PowerPoint PPT Presentation

Chapter 15. Applying equilibrium. The Common Ion Effect. When the salt with the anion of a weak acid is added to that acid, Lowers the percent dissociation of the acid. Ex: NaF and HF mixed together F- is the common ion HF ↔ H + + F -

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### Chapter 15

Applying equilibrium

The Common Ion Effect
• When the salt with the anion of a weak acid is added to that acid,
• Lowers the percent dissociation of the acid.
• Ex: NaF and HF mixed together
• F- is the common ion

HF ↔ H+ + F-

Adding F- shifts the equilibrium to the left and therefore fewer H+ ions present

The calculations are the same as last chapter.

NH4Cl and NH3 (NH4+ is the common ion)

NH3 + H+ NH4+

Buffered solutions
• A solution that resists a change in pH.
• Consists of either a weak acid and its salt or a weak base and its salt.
• We can make a buffer of any pH by varying the concentrations of these solutions.
• Same calculations as before.
• Calculate the pH of a solution that is .50 M HAc and .25 M NaAc (Ka = 1.8 x 10-5)

Na+ is a spectator and the reaction we are worried about is

HAc H+ + Ac-

• Choose x to be small

Initial 0.50 M 0 0.25 M

-x

x

x

Change

Final

0.50-x

x

0.25+x

• We can fill in the table

HAc H+ + Ac-

Initial 0.50 M 0 0.25 M

• Do the math
• Ka = 1.8 x 10-5

Change

-x

x

x

Final

0.50-x

x

0.25+x

x (0.25)

x (0.25+x)

=

1.8 x 10-5 =

(0.50)

(0.50-x)

• Assume x is small

x = 3.6 x 10-5

• Assumption is valid
• pH = -log (3.6 x 10-5) = 4.44
Adding a strong acid or base
• Do the stoichiometry first.
• Use moles not molar
• A strong base will grab protons from the weak acid reducing [HA]0
• A strong acid will add its proton to the anion of the salt reducing [A-]0
• Then do the equilibrium problem.
• What is the pH of 1.0 L of the previous solution when 0.010 mol of solid NaOH is added?

HAc H+ + Ac-

0.25 mol

0.50 mol

• In the initial mixture M x L = mol
• 0.50 M HAc x 1.0 L = 0.50 mol HAc

0.26 mol

0.49 mol

• 0.25 M Ac- x 1.0 L = 0.25 mol Ac-
• Adding 0.010 mol OH- will reduce the HAc and increase the Ac- by 0.010 mole
• Because it is in 1.0 L, we can convert it to molarity

HAc H+ + Ac-

0.25 mol

0.50 mol

0.26 M

0.49 M

• In the initial mixture M x L = mol
• 0.50 M HAc x 1.0 L = 0.50 mol HAc
• 0.25 M Ac- x 1.0 L = 0.25 mol Ac-
• Adding 0.010 mol OH- will reduce the HAc and increase the Ac- by 0.010 mole
• Because it is in 1.0 L, we can convert it to molarity

HAc H+ + Ac-

0.25 mol

0.50 mol

0.26 M

0.49 M

• Fill in the table

HAc H+ + Ac-

Initial 0.49 M 0 0.26 M

-x

x

x

Final

0.49-x

x

0.26+x

HAc H+ + Ac-

Initial 0.49 M 0 0.26 M

• Do the math
• Ka = 1.8 x 10-5

-x

x

x

Final

0.49-x

x

0.26+x

x (0.26)

x (0.26+x)

=

1.8 x 10-5 =

(0.49)

(0.49-x)

• Assume x is small

x = 3.4 x 10-5

• Assumption is valid
• pH = -log (3.4 x 10-5) = 4.47
Notice
• If we had added 0.010 mol of NaOH to 1 L of water, the pH would have been.
• 0.010 M OH-
• pOH = 2
• pH = 12
• But with a mixture of an acid and its conjugate base the pH doesn’t change much
• Called a buffer.
General equation
• Ka = [H+] [A-] [HA]
• so [H+] = Ka [HA] [A-]
• The [H+] depends on the ratio [HA]/[A-]
• taking the negative log of both sides
• pH = -log(Ka [HA]/[A-])
• pH = -log(Ka)-log([HA]/[A-])
• pH = pKa + log([A-]/[HA])
This is called the Henderson-Hasselbach equation
• pH = pKa + log([A-]/[HA])
• pH = pKa + log(base/acid)
• Works for an acid and its salt
• Like HNO2 and NaNO2
• Or a base and its salt
• Like NH3 and NH4Cl
• But remember to change Kb to Ka
Calculate the pH of the following
• 0.75 M lactic acid (HC3H5O3) and 0.25 M sodium lactate (Ka = 1.4 x 10-4)
• pH = 3.38
Calculate the pH of the following
• 0.25 M NH3 and 0.40 M NH4Cl (Kb = 1.8 x 10-5)
• Ka = 1 x 10-14 1.8 x 10-5
• Ka = 5.6 x 10-10
• remember its the ratio base over acid
• pH = 9.05
Prove they’re buffers
• What would the pH be if .020 mol of HCl is added to 1.0 L of both of the preceding solutions.
• What would the pH be if 0.050 mol of solid NaOH is added to 1.0 L of each of the proceeding.
• Remember adding acids increases the acid side,
• Adding base increases the base side.
Prove they’re buffers
• What would the pH be if .020 mol of HCl is added to 1.0 L of preceding solutions.
• 0.75 M lactic acid (HC3H5O3) and 0.25 M sodium lactate (Ka = 1.4 x 10-4)

HC3H5O3 H+ + C3H5O3-

Initially 0.75 mol 0 0.25 mol

After acid 0.77 mol 0.23 mol

Compared to 3.38 before acid was added

Prove they’re buffers
• What would the pH be if 0.050 mol of solid NaOH is added to 1.0 L of the solutions.
• 0.75 M lactic acid (HC3H5O3) and 0.25 M sodium lactate (Ka = 1.4 x 10-4)

HC3H5O3 H+ + C3H5O3-

Initially 0.75 mol 0 0.25 mol

After acid 0.70 mol 0.30 mol

Compared to 3.38 before acid was added

Prove they’re buffers
• What would the pH be if .020 mol of HCl is added to 1.0 L of preceding solutions.
• 0.25 M NH3 and 0.40 M NH4Cl Kb = 1.8 x 10-5 so Ka = 5.6 x 10-10

NH4+H+ + NH3

Initially 0.40 mol 0 0.25 mol

After acid 0.42 mol 0.23 mol

Compared to 9.05 before acid was added

Prove they’re buffers
• What would the pH be if 0.050 mol of solid NaOH is added to 1.0 L each solutions.
• 0.25 M NH3 and 0.40 M NH4Cl Kb = 1.8 x 10-5 so Ka = 5.6 x 10-10

NH4+H+ + NH3

Initially 0.40 mol 0 0.25 mol

After acid 0.35 mol 0.30 mol

Compared to 9.05 before acid was added

Buffer capacity
• The pH of a buffered solution is determined by the ratio [A-]/[HA].
• As long as this doesn’t change much the pH won’t change much.
• The more concentrated these two are the more H+ and OH- the solution will be able to absorb.
• Larger concentrations = bigger buffer capacity.
Buffer Capacity
• Calculate the change in pH that occurs when 0.040 mol of HCl(g) is added to 1.0 L of each of the following:
• 5.00 M HAc and 5.00 M NaAc
• 0.050 M HAc and 0.050 M NaAc
• Ka= 1.8x10-5
• pH = pKa
Buffer Capacity
• Calculate the change in pH that occurs when 0.040 mol of HCl(g) is added to 1.0 L of 5.00 M HAc and 5.00 M NaAc
• Ka= 1.8x10-5

HAc H+ + Ac-

Initially 5.00 mol 0 5.00 mol

After acid 5.04 mol 4.96 mol

Compared to 4.74 before acid was added

Buffer Capacity
• Calculate the change in pH that occurs when 0.040 mol of HCl(g) is added to 1.0 L of 0.050 M HAc and 0.050 M NaAc
• Ka= 1.8x10-5

HAc H+ + Ac-

Initially 0.050 mol 0 0.050 mol

After acid 0.090 mol 0 0.010 mol

Compared to 4.74 before acid was added

Buffer capacity
• The best buffers have a ratio [A-]/[HA] = 1
• This is most resistant to change
• True when [A-] = [HA]
• Makes pH = pKa (since log 1 = 0)

### Titrations

Titrations
• Millimole (mmol) = 1/1000 mol
• Molarity = mmol/mL = mol/L
• Makes calculations easier because we will rarely add liters of solution.
• Adding a solution of known concentration until the substance being tested is consumed.
• This is called the equivalence point.
• Where moles of acid = moles of base
Strong acid with strong Base
• Equivalence at pH 7

7

pH

Weak acid with strong Base

• Equivalence at pH >7

>7

7

pH

Strong base with strong acid

• Equivalence at pH 7

7

pH

Weak base with strong acid

• Equivalence at pH <7

7

<7

pH

Strong acid with Strong Base
• Do the stoichiometry.
• mL x M = mmol
• There is no equilibrium .
• They both dissociate completely.
• The reaction is H+ + OH- HOH
• Use [H+] or [OH-] to figure pH or pOH
• The titration of 20.0 mL of 0.10 M HNO3 with 0.10 M NaOH
Weak acid with Strong base
• There is an equilibrium.
• Do stoichiometry.
• Use moles
• Determine major species
• Then do equilibrium.

Calculate the pH after 0.0mL, 15.0mL, 25.0 mL, and 30.0mL of 0.10 M NaOH are added to 25.0mL of 0.10M of Nicotinic acid Ka= 1.4 x 10 -5

Summary
• Strong acid and base just stoichiometry.
• Weak acid with 0 ml of base - Ka
• Weak acid before equivalence point
• Stoichiometry first
• Then Henderson-Hasselbach
• Weak acid at equivalence point- Kb -Calculate concentration
• Weak acid after equivalence - leftover strong base. -Calculate concentration
Example
• A 25.0 mL sample of benzoic acid (0.120M) is titrated w/ NaOH (.210 M) Ka= 6.3 x 10 -5
• Calculate # moles of Benzoic acid
• Calculate volume of NaOH needed to reach equiv. pt
• Calc. the pH before any base is added
• Calc. pH after 10ml base is added
• Find the pH at equivalence pt
• Calculate the pH after 15.3mL of OH- is added
Summary
• Weak base before equivalence point.
• Stoichiometry first
• Then Henderson-Hasselbach
• Weak base at equivalence point Ka. -Calculate concentration
• Weak base after equivalence – left over strong acid. -Calculate concentration
Indicators
• Weak acids that change color when they become bases.
• weak acid written HIn
• Weak base
• HIn H+ + In-clear red
• Equilibrium is controlled by pH
• End point - when the indicator changes color.
• Try to match the equivalence point
Indicators
• Since it is an equilibrium the color change is gradual.
• It is noticeable when the ratio of [In-]/[HI] is 1/10 or [HI]/[In-] is 10/1
• Since the Indicator is a weak acid, it has a Ka.
• pH the indicator changes is
• pH=pKa +log([In-]/[HI]) = pKa +log(1/10)
• pH=pKa - 1
Indicators
• pH=pKa + log([HI]/[In-]) = pKa + log(10)
• pH=pKa+1
• Choose the indicator with a pKa 1 more than the pH at equivalence point if you are titrating with base.
• Choose the indicator with a pKa 1 less than the pH at equivalence point if you are titrating with acid.
Sample Problem
• Two drops of indicator HIn (Ka= 1.0 x 10 -9), hwere HIn is yellow and In- is blue are placed in 100.0 mL of 0.10 M HCl.
• What color is the solution initially?
• The solution is titrated with 0.10 M NaOH. At what pH will the color change (greenish yellow) occur?
• What color will the solution be after 200.0mL of NaOH has been added?

### Solubility Equilibria

Will it all dissolve, and if not, how much?

All dissolving is an equilibrium.
• If there is not much solid it will all dissolve.
• As more solid is added the solution will become saturated.
• Solid dissolved
• The solid will precipitate as fast as it dissolves .
• Equilibrium
General equation
• M+ stands for the cation (usually metal).
• Nm-stands for the anion (a nonmetal).
• MaNmb(s) aM+(aq) + bNm- (aq)
• Remember the concentration of a solid doesn’t change. So, what’s the K expression?
• Ksp = [M+]a[Nm-]b
• Called the solubility product for each compound.
Watch out
• Solubility is not the same as solubility product.
• Solubility product is an equilibrium constant.
• it doesn’t change except with temperature.
• Solubility is an equilibrium position-usually expressed as a Molarity for how much can dissolve.
Calculating Solubility

Ex: Calc. Solubility of CaC2O4 if the value of Ksp= 4.8 x 10 -5 mol/L

CaC2O4 (s) ↔ Ca+2(aq) + C2O4 -2 (aq)

Ksp = [Ca+2] [C2O4-2]

Ksp= [s] [s] = [s]2 = 4.8x10-5 = s2

s= 6.9 x 10 -3 M

Calculating Ksp
• The solubility of copper(I) bromide is

2.0 x 10 -4 mol/L. Calc. Ksp value.

CuBr(s) ↔ Cu+1(aq) + Br-1 (aq)

I .0002M 0M 0M

C - .0002M +.0002M +.0002M

E 0 .0002M .0002M

Ksp = [.0002] [.0002] = 4.0 x 10 -8

(units usually omitted)

Calculating Ksp
• The solubility of iron(II) oxalate FeC2O4 is 65.9 mg/L
• The solubility of Li2CO3 is 5.48 g/L
Common Ion Effect
• If we try to dissolve the solid in a solution with either the cation or anion already present less will dissolve.
• Calculate the solubility of SrSO4, with a Ksp of 3.2 x 10-7 in M in a solution of 0.010 M Na2SO4.
• Calculate the solubility of CaF2

Ksp= 4.0 x 10 -11 in a 0.025 M NaF solution.

AgCl and NaCl in solution
• Cl- is the common ion
• When [Ag+] [Cl-] < Ksp, no precipitate will be formed.
• When [Ag+] [Cl-] > Ksp, a precipitate will be formed.
pH and solubility
• OH- can be a common ion.
• More soluble in acid.
• For other anions if they come from a weak acid they are more soluble in a acidic solution than in water.
• CaC2O4 Ca+2 + C2O4-2
• H+ + C2O4-2 HC2O4-
• Reduces [C2O4-2] in acidic solution.
Sample problem
• What is the pH in a saturated solution of Ca(OH)2?Ksp = 5.5 x 10-6 for Ca(OH)2.
Q and Precipitation
• not necessarily at Equil.
• Q= Ion product
• Q =[M+]a[Nm-]b
• If Q>Ksp a precipitate forms. (supersaturated)
• If Q<Ksp No precipitate. (unsaturated)
• If Q = Ksp equilibrium. (saturated)
Example1

Chemical analysis gave [Pb2+] = 0.012 M, and [Br-] = 0.024 M in a solution. From a table, you find Ksp for PbBr2 has a value of 4x 10-5. Is the solution saturated, oversaturated or unsaturated?

Example 2
• A solution of 750.0 mL of 4.00 x 10-3M Ce(NO3)3 is added to 300.0 mL of 2.00 x 10-2M KIO3. Will Ce(IO3)3 (Ksp= 1.9 x 10-10M) precipitate and if so, what is the concentration of the ions?
Selective Precipitations
• Used to separate mixtures of metal ions in solutions.
• Add anions that will only precipitate certain metals at a time.
• Used to purify mixtures.
• Often use H2S because in acidic solution Hg+2, Cd+2, Bi+3, Cu+2, Sn+4 will precipitate.
Selective Precipitation
• Then add OH-solution [S-2] will increase so more soluble sulfides will precipitate.
• Co+2, Zn+2, Mn+2, Ni+2, Fe+2, Cr(OH)3, Al(OH)3
Selective precipitation
• First with insoluble chlorides (Ag, Pb, Ba)
• Then sulfides in Acid.
• Then sulfides in base.
• Then insoluble carbonate (Ca, Ba, Mg)
• Alkali metals and NH4+ remain in solution.
Complex ion Equilibria
• A charged ion surrounded by ligands.
• Ligands are Lewis bases using their lone pair to stabilize the charged metal ions.
• Common ligands are NH3, H2O, Cl-,CN-
• Coordination number is the number of attached ligands.
• Cu(NH3)42+ has a coordination # of 4
The addition of each ligand has its own equilibrium
• Usually the ligand is in large excess.
• And the individual K’s will be large so we can treat them as if they go to completion.
• The complex ion will be the biggest ion in solution.
Calculate the concentrations of Ag+, Ag(S2O3)-2, and Ag(S2O3)2-3 in a solution made by mixing 150.0 mL of 0.010 M AgNO3 with 200.0 mL of 5.00 M Na2S2O3
• Ag+ + S2O3-2 Ag(S2O3)- K1=7.4 x 108
• Ag(S2O3)- + 2S2O3-2 Ag(S2O3)2-3 K2=3.9 x 104

(Hint: set up ice problem, determine L.R.)

This problem is found on pg. 768-769 in your book. Check your work!

Complex Ions and Solubility

The dissolving of solid AgCl in excess NH3 is represented as…

AgCl(s) + 2NH3(aq) ↔ Ag(NH3)2+ (aq) + Cl-(aq)

However, this is a result of several steps

AgCl(s) ↔ Ag+ + Cl- Ksp= 1.6 x 10-10

Ag+ + NH3 ↔ Ag(NH3)+ K1 = 2.1 x 103

Ag(NH3)+ + NH3 ↔ Ag(NH3)2+ K2= 8.2 x 103

AgCl(s) + 2NH3(aq) ↔ Ag(NH3)2+ (aq) + Cl-(aq)

So the Equilibrium Expression is written as…. K= [Ag(NH3)2+] [Cl-]

[NH3]2

= Ksp x K1 x K2

= (1.6 x 10-10) ( 2.1 x 103) (8.2 x 103)

= 2.8 x 10-3

Summary

3 Factors that affect solubility:

- Common Ions

- pH

- Complexing Agents/Ions

Mixtures of Ions can be separated by selective precipitation (you know this as Quantitative Analysis)

- Steps listed on pg.765 or in your lab